Ask your own question, for FREE!
Mathematics 19 Online
OpenStudy (anonymous):

f(x) = tan(x) ^ arctan(x) Find f'(x).

OpenStudy (anonymous):

\[f(x) = (\tan x)^{\arctan x}\]

OpenStudy (anonymous):

believe the only function that acts as its own inverse (i.e. f^-1(x) = f(x)) is y = x.

OpenStudy (anonymous):

did i help at all?

OpenStudy (anonymous):

not really, the answer in the back of the book is ridiculously long

OpenStudy (baldymcgee6):

http://gyazo.com/83c8c62b3f1098672c155d554aee7da2

OpenStudy (anonymous):

u have to take logarithm of both sides and then u get ln(f(x))=arctan(x)ln(tanx) is it f'(x)= tan(x) ^ arctan(x)((ln(tanx)/(1+x^2)) + (arctanx)((sec^2x)/(tanx)))) =tan(x) ^ arctan(x)( ln(tanx)/(1+x^2) + (arctanx)(secx)(cosecx) )

OpenStudy (anonymous):

The final answer anded up being \[f'(x) = tanx ^{arctanx}\left( \frac{ lntanx }{ 1+x^2 } + arctanxcotxsec^2x\right)\] Thanks!

OpenStudy (anonymous):

welcome i hope u understand that (arctanx)(secx)(cosecx) =(arctanx)(sec^2x)(cotx)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!